cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A186500 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2. Complement of A186499.

Original entry on oeis.org

2, 6, 9, 12, 16, 19, 22, 25, 29, 32, 35, 38, 42, 45, 48, 51, 54, 58, 61, 64, 67, 71, 74, 77, 80, 84, 87, 90, 93, 97, 100, 103, 106, 110, 113, 116, 119, 122, 126, 129, 132, 135, 139, 142, 145, 148, 152, 155, 158, 161, 165, 168, 171, 174, 177, 181, 184, 187, 190, 194, 197, 200, 203, 207, 210, 213, 216, 220, 223, 226, 229, 232, 236, 239, 242, 245, 249, 252, 255, 258, 262, 265
Offset: 1

Views

Author

Clark Kimberling, Feb 22 2011

Keywords

Comments

See A186219 for a discussion of adjusted joint rank sequences.
The pairs (i,j) for which i^2=-4+5j^2 are (L(2h-2),F(2h-1)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers).

Examples

			First, write
1..4..9..16..25..36..49..... (i^2)
1........16........41........(-4+5j^2)
Then replace each number by its rank, where ties are settled by ranking i^2 before -4+5j^2:
a=(1,3,4,5,7,8,10,11,13,14,15,17,18...)=A186499
b=(2,6,9,12,16,19,22,25,29,32,35,38,.)=A186500.
		

Crossrefs

Programs

Formula

a(n)=n+floor((1/10)(sqrt(2n^2+7)))=A186499(n).
b(n)=n+floor(sqrt(5n^2-7/2))=A186500(n).

A186511 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2. Complement of A186512.

Original entry on oeis.org

2, 3, 4, 6, 7, 8, 10, 11, 13, 14, 16, 17, 18, 20, 21, 23, 24, 26, 27, 28, 30, 31, 33, 34, 36, 37, 39, 40, 42, 43, 44, 46, 47, 49, 50, 52, 53, 55, 56, 57, 59, 60, 62, 63, 65, 66, 68, 69, 70, 72, 73, 75, 76, 78, 79, 81, 82, 83, 85, 86, 88, 89, 91, 92, 94, 95, 96, 98, 99, 101, 102, 104, 105, 107, 108, 110, 111, 112, 114, 115, 117, 118, 120, 121, 123, 124, 125, 127, 128, 130, 131, 133, 134, 136, 137, 138, 140, 141, 143, 144
Offset: 1

Views

Author

Clark Kimberling, Feb 22 2011

Keywords

Comments

See A186219 for a discussion of adjusted joint rank sequences.
The pairs (i,j) for which i^2=-4+5j^2 are (L(2h-1),F(2h-1)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers).

Examples

			First, write
1..4..9..16..25..36..49..... (i^2)
1........16........41........(-4+5j^2)
Then replace each number by its rank, where ties are settled by ranking i^2 after -4+5j^2:
a=(2,3,4,6,7,8,10,11,13,14,16,...)=A186511
b=(1,5,9,12,15,19,22,25,29,32,...)=A186512.
		

Crossrefs

Programs

  • Mathematica
    (* adjusted joint rank sequences a and b, using general formula for ranking ui^2+vi+w and xj^2+yj+z *)
    d = -1/2; u = 1; v = 0; w = 0; x = 5; y = 0; z =-4;
    h[n_] := -y + (4 x (u*n^2 + v*n + w - z - d) + y^2)^(1/2);
    a[n_] := n + Floor[h[n]/(2 x)];
    k[n_] := -v + (4 u (x*n^2 + y*n + z - w + d) + v^2)^(1/2);
    b[n_] := n + Floor[k[n]/(2 u)];
    Table[a[n], {n, 1, 100}]  (* A186511 *)
    Table[b[n], {n, 1, 100}]  (* A186512 *)

Formula

a(n)=n+floor(sqrt((n^2)/5+9/10))=A186511(n).
b(n)=n+floor(sqrt(5n^2-9/2))=A186512(n).

A186512 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) after g(j) when f(i)=g(j), where f(i)=i^2 and g(j)=-4+5j^2. Complement of A186511.

Original entry on oeis.org

1, 5, 9, 12, 15, 19, 22, 25, 29, 32, 35, 38, 41, 45, 48, 51, 54, 58, 61, 64, 67, 71, 74, 77, 80, 84, 87, 90, 93, 97, 100, 103, 106, 109, 113, 116, 119, 122, 126, 129, 132, 135, 139, 142, 145, 148, 152, 155, 158, 161, 165, 168, 171, 174, 177, 181, 184, 187, 190, 194, 197, 200, 203, 207, 210, 213, 216, 220, 223, 226, 229, 232, 236, 239, 242, 245, 249, 252, 255, 258, 262, 265
Offset: 1

Views

Author

Clark Kimberling, Feb 22 2011

Keywords

Comments

See A186219 for a discussion of adjusted joint rank sequences.
The pairs (i,j) for which i^2=-4+5j^2 are (L(2h-1),F(2h-1)), where L=A000032 (Lucas numbers) and F=A000045 (Fibonacci numbers).

Examples

			First, write
1..4..9..16..25..36..49..... (i^2)
1........16........41........(-4+5j^2)
Then replace each number by its rank, where ties are settled by ranking i^2 after -4+5j^2:
a=(2,3,4,6,7,8,10,11,13,14,16,...)=A186511
b=(1,5,9,12,15,19,22,25,29,32,...)=A186512.
		

Crossrefs

Programs

Formula

a(n)=n+floor(sqrt((n^2)/5+9/10))=A186511(n).
b(n)=n+floor(sqrt(5n^2-9/2))=A186512(n).

A186497 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=3i-2 and g(j)=j-th triangular number. Complement of A186498.

Original entry on oeis.org

1, 4, 6, 7, 9, 11, 12, 14, 15, 16, 18, 19, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 34, 35, 36, 37, 39, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 54, 56, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 71, 72, 73, 74, 76, 77, 78, 79, 80, 81, 82, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 99, 100, 101, 102, 103, 104, 105, 106, 108, 109, 110, 111, 112, 113, 114, 116, 117, 118, 119, 120, 121, 122, 123, 125, 126, 127, 128, 129, 130, 131, 132, 133, 135, 136, 137, 138, 139, 140, 141, 142, 144, 145, 146
Offset: 1

Views

Author

Clark Kimberling, Feb 22 2011

Keywords

Comments

See A186350 for a discussion of adjusted joint rank sequences.

Examples

			First, write
1..4..7.10..13..16..19..22..25..28..31. (3i-2),
1.3..6..10....15.......21.......28.....(j(j+1)/2).
Then replace each number by its rank, where ties are settled by ranking 3i-2 before j(j+1)/2:
a=(1,4,6,7,9,11,12,14,15,16,18,...)=A186497,
b=(2,3,5,8,10,13,17,20,24,29,33,..)=A186498.
		

Crossrefs

Programs

  • Mathematica
    (* Adjusted joint rank sequences a and b, using general formula for ranking 1st degree u*n+v and 2nd degree x*n^2+y*n+z. *)
    d=1/2; u=3; v=-2; x=1/2; y=1/2;
    h[n_]:=(-y+(4x(u*n+v-d)+y^2)^(1/2))/(2x);
    a[n_]:=n+Floor[h[n]];
    k[n_]:=(x*n^2+y*n-v+d)/u;
    b[n_]:=n+Floor[k[n]];
    Table[a[n],{n,1,120}]  (* A186497 *)
    Table[b[n],{n,1,100}]  (* A186498 *)

A186498 Adjusted joint rank sequence of (f(i)) and (g(j)) with f(i) before g(j) when f(i)=g(j), where f(i)=3i-2 and g(j)=j-th triangular number. Complement of A186497.

Original entry on oeis.org

2, 3, 5, 8, 10, 13, 17, 20, 24, 29, 33, 38, 44, 49, 55, 62, 68, 75, 83, 90, 98, 107, 115, 124, 134, 143, 153, 164, 174, 185, 197, 208, 220, 233, 245, 258, 272, 285, 299, 314, 328, 343, 359, 374, 390, 407, 423, 440, 458, 475, 493, 512, 530, 549, 569, 588, 608, 629, 649, 670, 692, 713, 735, 758, 780, 803, 827, 850, 874, 899, 923, 948, 974, 999, 1025, 1052, 1078, 1105, 1133, 1160, 1188, 1217, 1245, 1274, 1304
Offset: 1

Views

Author

Clark Kimberling, Feb 22 2011

Keywords

Examples

			First, write
1..4..7.10..13..16..19..22..25..28..31. (3i-2),
1.3..6..10....15.......21.......28.....(j(j+1)/2).
Then replace each number by its rank, where ties are settled by ranking 3i-2 before j(j+1)/2:
a=(1,4,6,7,9,11,12,14,15,16,18,...)=A186497,
b=(2,3,5,8,10,13,17,20,24,29,33,..)=A186498.
		

Crossrefs

Programs

Showing 1-5 of 5 results.